Rolling friction is linear in both velocity and vehicle mass, and has a CR (rolling resistance coefficient) of about 0.0055. It is proportional to the force (Newtons) normal to the track. See http://en.wikipedia.org/wiki/Rolling_resistance. For all four wheels, we get;

FR = -CRmg cos(θ)

Ft is pretty straightforward - for a cartie of mass m accelerating at a;

Ft = m a

The Fw term is a little more fiddly; the angular acceleration ω of the wheel with moment of inertia I is caused by the torque T

T = I ω

I for a hoop of mass mw is mw r2, and ω is given by a / r, so the above equation becomes;

T = mw r2 (a / r) T = mw r a

T also equals Fw r, so

Fw r = mw r aFw = mw a

i.e. The radius of the wheel has no effect - the extra inertia of the larger wheel is cancelled out by the larger torque applied to rotate it.

Substituting these two terms back into the initial equation we now have;

The equations above can be rearranged to calculate the maximum speed of a soapbox on any given slope (provided the course is long enough).

Maximum speed is achieved when the force on the cartie due to gravity is equal to the forces due to rolling resistance, moment of inertia and drag. i.e;

Ft = Fd + FR + Fw

so

Fd = Ft - FR - Fw

(1/2)ρACdv2 = = Ft - FR- Fw

or, rearranging for V,

V = √ (2(Ft - FR - Fw) / ρCdA)

Substituting the equations above in for the forces gives;

V = √ (2(m g sin(θ) - CRm g cos(θ) - mw a) / ρCdA)

But when V is at the maximum, acceleration a = 0, so the mw a term disappears;

V = √ (2(m g sin(θ) - CRm g cos(θ) ) / ρCdA)

So. That's all very clever, but what does it mean? Well - if you plug in some typical values for the variables, you can plot a set of curves to show how mass and drag affect your terminal velocity.

For instance,on a slope of gradient 1 in 10 (10%), with a rolling resistance of .0055, air at standard temperature and pressure, you get the following graph;

I.e. Heavier carties go faster, but the increase in top speed per Kg decreases as overall weight increases. And it doesn't really matter how much weight you chuck into your cartie if it has the aerodynamics of a warehouse as you'll just never catch a slick and streamlined machine no matter how hard you try.

The other interesting thing that drops out of this analysis is the slightly surprising observation that the mass and diameter of your wheels has no effect on your terminal velocity.

Well - sort of. There are still frictional losses through tyre deformation, bearings etc, and of course your wheels have aerodynamic drag too. But when you consider your wheels in terms of the force used to increase the speed at which they are rotating, at terminal velocitythere is no force required to do this.